# Curved space puzzle

1. Apr 14, 2010

### Nickelodeon

If you think of a bow (as in bow and arrow) placed in space in the proximity of a massive object. This bow for arguments sake is 100 km from tip to tip.

Now replace the bow (wooden part and string) with what might be considered waypoints. The curved wooden part represents a straight line in curved space and the taught string would, I suppose, appear curved.

If one person sets off along the 'wooden' path and simultaneously someone else sets off along the 'string' path what would be the outcome

a) They would both arrive at the furthermost tip at the same time but the person who chose the string route was older?

b) The person who chose the 'wooden' path arrived first?

c) The person who chose the 'string' path arrived first?

d) They both arrived together and their biological clocks where still in sync?

e) The string route was shorter in distance but required more energy to traverse?

f) none of the above?

I'm not sure but I would think answer 'e' seems likely and probably 'c'

2. Apr 14, 2010

### starthaus

The problem as stated is ill-posed. This is a variant of the twins paradox set in a gravitational field. Whether or not the twins arrive at destination at the same time is obviously a function of their respective speeds and their respective path lengths. If they do not arrive simultneously, the problem is tougher.
Let's assume that they twins arrive simultaneously, at time "T". Their total elapsed proper times is can be calculated from the Schwarzschild metric, via integration wrt the coordinate time. If the massive gravitational object is rotating, you will need to replace the Schwarzschild metric with the Kerr one. Here is a sketch of the solution:

$$d\tau^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2 (d \phi)^2$$

$$d\tau=dt \sqrt(\alpha -\frac{1}{\alpha}\frac{dr^2}{dt^2}-r^2 \frac{d \phi^2}{dt^2})$$

Integrate the above wrt "t" with 0<t<T and you have your solution. You will need to note that for the twin travelling along the bow chord

$$d \phi=0$$.

For the twin travelling along the bow, you have

$$\phi=f(r)$$

Nasty stuff :-) Things get even nastier since

$$\alpha=1-2m/r$$

Last edited: Apr 14, 2010
3. Apr 15, 2010

### Nickelodeon

What I was trying to get straight in my mind was whether in curved space it is quicker to reach a destination by not taking what would appear to be the straight line path, ie take a short cut. For instance, in an extreme situation where the curvature forces light to veer 90 degrees then could one assume that the shortest time to a destination would be by not following the more obvious route. (in my example above the more obvious route would be along the wooded bit of the bow) - probably not explaining myself very well.